Poj 2516 Minimum Cost
Link: http://poj.org/problem? Id = 2516
Question:There are m warehouses, N merchants, and K items. Input n, m, K first. Then, enter the number of K in N rows. Each row represents the items to be purchased by a merchant. K numbers represent the number of items to be purchased. Then there is the k Number in m rows. Each row indicates the situation in the warehouse, where k number indicates the inventory of each item. Next, there
. the first line of each test case contains three integers n, m, K (0
Then come K integer matrices (each with the size N * m), The INTEGER (this integer is belong to (0,100) at the I-th row, j-th Column in the k-th matrix represents the cost to transport one unit of k-th goods from the J-th supply place to the I-th shopkeeper.
The input is terminated with three "0" S. This test case shocould not be processed.
Output For each test case, if dear
Orz Creator zkw%%%Send the zkw god Benniu's blog in situ: Portal
The cost stream is based on the maximum flow of the network, the largest stream in a graph has and has only one, but the maximum number of stream bars is often more than one, and for us it may be necessary to find the smallest (or largest) path in these m
Question address: POJ2195
I am the first to get my career expenses !! After the invitational competition, I decided to learn more. At least I had to drop A simple network stream. In the future, the maximum flow cost will be reduced to a single click.
In the past, the cost flow had always been a tall and unattainable image in my mind. But in the past two days, I f
Question address: poj2195
I am the first to get my career expenses !! After the invitational competition, I decided to learn more. At least I had to drop a simple network stream. In the future, the maximum flow cost will be reduced to a single click.
In the past, the cost flow had always been a tall and unattainable image in my mind. But in the past two days, I f
", S,ODD[I],B[ODD[I]],0LL)*/; for(intI=1; i/*, printf ("%d-->%d:%d/%i64d\n", EVEN[I],T,B[EVEN[I]],0LL)*/; //printf ("%d%d%d\n", ot,et,cnt);}intMain () {//freopen ("menci_pair.in", "R", stdin);//freopen ("Menci_pair.out", "w", stdout);n=read (); for(intI=1; iread (); for(intI=1; iread (); for(intI=1; iLong Long) read (); Build (); while(SPFA () !f) maxcostflow (); printf ("%d\n", Flow); return 0;}At that time the only thought and the problem of the same solution, but finally did not know
HDU 3488 -- Tour [minimum cost maximum stream classic], hdu3488 -- tour
Tour
Time Limit: 3000/1000 MS (Java/Others) Memory Limit: 65535/65535 K (Java/Others)Total Submission (s): 2308 Accepted Submission (s): 1156
Problem DescriptionIn the kingdom of Henryy, there are N (2 Every city shoshould be just in one route.
A loop shoshould have at least two cities. in one route, each city shoshould be visited ju
Matrix (maximum cost of hdu 2686 Stream)
Matrix
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)Total Submission (s): 1893 Accepted Submission (s): 1006
Problem Description Yifenfei very like play a number game in the n * n Matrix. A positive integer number is put in each area of the Matrix.
Every time yifenfei shocould to do is that choose a detour which frome the top left
system by backtracking root analysis (RCA);3, the cost control line , mainly budget financial management, the formation of a project-based budget and accounting mechanism has a certain degree of difficulty, and even need to revise our reimbursement system, each cost should be clearly implemented to a specific project to bear, how many projects this year, how much each project
Question link: http://acm.hdu.edu.cn/showproblem.php? PID = 1, 5988
Question:
Given N points and M directed edges, each point is a place to eat and each person has a box of meals. There are s people at each point, and there are B lunch boxes. Each side can only be walked C times, and each side has a wire,When the first person passes, the wire will not be damaged. From the second person, each time there is a probability that P will destroy the wire. So that everyone can eat and find the minimum p
Label: style blog HTTP color Io OS AR for SP
Hmm ~ The first write cost flow question...
This is the template Question of the billing flow. No more bare questions can be found.
Graph creation: Each m (man) serves as the source point, each H (house) serves as the sink point, and each source point and sink point connects one edge separately, the traffic on this edge is 1 (because each source point can only take one edge to the sink point), and the
HDU 2686 Matrix (maximum cost Stream)MatrixTime Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)Total Submission (s): 1890 Accepted Submission (s): 1005Problem Description Yifenfei very like play a number game in the n * n Matrix. A positive integer number is put in each area of the Matrix.Every time yifenfei shocould to do is that choose a detour which frome the top left point to
HDU 2686 Matrix 3376 Matrix Again (cost stream), hdu3376HDU 2686 Matrix
Question Link3376 Matrix Again
Question Link
These two questions are the same, but the data ranges are different. They are all a matrix. You can get the maximum value from the upper left corner to the lower right corner to the upper left corner.
Idea: Split the points, create a graph, and then run the fee
HDU 5045 Contest (cost stream), hdu5045
Address: HDU 5045
Finally, I used network stream A in the competition... I used so many network streams once .. Although the question is very simple, I still want to remember it...
My idea for this question is to find m/n charge streams, and calculate the maximum charge flow for the same round for every n of them. Creating
HDU 3315 My Brute (cost Stream)
Address: HDU 3315
The idea of this question is completely self-developed, and I feel quite clever... (Do not spray the ox ...) I have more confidence in building bold images.
The specific idea is to construct a bipartite graph. The Si source and Xi connection points have 1 traffic and the cost is 0. then, when Si can win Xj, the tw
HDU 4862 Jump (Multi-school joint training 1) (minimum cost maximum stream), hdu4862
Address: HDU4862
The minimum cost flow is still too small. You cannot figure it out...
Directly reference the official question...
The minimum K path coverage model is solved by the cost flow or KM algorithm. A two-part graph is constr
/* Question: Given a n * m map, there are several man and house on the map, and the number of man and house are consistent. $1 (unit cost = unit distance) is required for a man to move one cell. Only one man can be accommodated in one house. Now all man is required to stay in the house and the minimum fee is required. The minimum cost of the maximum flow to the spfa to find the minimum
Label: Minimum fee of network flow, maximum flow, algorithm, Bipartite Graph
Title: hdoj 3488 tour
Question: give n vertices and m edges, and then ask you to calculate that each vertex can only be in one ring (Hamilton ring), and the minimum cost of all vertices is only one time.
Analysis
Graph creation scheme:
Split each vertex into two I and II
Then s connects all I, capacity 1, cost 0
II connect all t,
"Test Instructions" gives an no-map, starting from 1 to N, two without a common edge of the shortest, so that the sum of the distance is minimal
The weight of each edge is set to cost, the maximum flow is set to 1, and then is the minimum charge flow from the source point to the meeting point flow rate of 2.Because it is the specified traffic, a new source point and a sink point, the source point to the node 1 a maximum flow of 2, the
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